Q: A particle is projected from the mid-point of the line joining two fixed particles each of mass m. If the separation between the fixed particles is l, the minimum velocity of projection of the particle so as to escape is equal to
(A)√(Gm/l)
(B)√(Gm/2l)
(C )√(2Gm/l)
(D) 2√(2Gm/l)

Q: Two massive particles of masses M & m (M > m) are separated by a distance l . They rotate with equal angular velocity under their gravitational attraction. The linear speed of the particle of mass m is
(A)√GMm/(M+m)l
(B) √GM²/(M+m)l
(C)√Gm/l
(D)√Gm²/(M+m)l
Solution : The system rotates about the centre of mass. The gravitational force acting on the particle m accelerates it towards the centre of the circular path, which has the radius
r = Ml /(M+m)
 F = mv²/r
GMm/l² = mv²/(Ml /(M+m)) By Solving ,

Q: A body is fired with velocity of magnitude √gR< V<√2gR at an angle of 30° with the radius vector of earth. If at the highest point the speed of the body is v/4, the maximum height attained by the body is equal to
(A) v²/8g
(B) R
(C) √2 R
(D) none of these
where R = radius of earth. (Use the appropriate assumptions)
Solution :
Conservation of angular momentum of the body about O yields
mv (sin30°) R = mv'(R + h)
 ½ vR = ¼ v(R + h) 
Hence , h = R.

Q: A satellite close to earth is in orbit above the equator with a period of rotation of 1.5 hrs. If it is above a point P on the equator at some time, it will be above P again after time
(A) 1.5 hrs.
(B) 1.6 hrs. if it is rotating from west to east.
(C) 24/17 hrs. if it is rotating from west to east.
(D) 24/17 hrs. if it is rotating from east to west.

Solution : Let ωo = the angular velocity of earth above its axis = 2π/24 rad/hr
Let ω = angular velocity of satellite.
  = 2π/1.5
For a satellite rotating from west to east. (same as earth), the relative angular velocity ω1 = ω – ωo
 Time period of rotation relative to earth =2π/ω1 = 1.6 h
Now for a satellite rotating from east to west (opposite to earth) the relative angular velocity ω2 = ω + ωo
Time period of rotation relative to earth =2π/ω2 = 24/17 hrs.